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Algebra Difficulty 4.9 AIME Find the answer

The polynomial x3−3x2+1x^{3}-3 x^{2}+1 has three real roots r1,r2r_{1}, r_{2}, and r3r_{3}. Compute 3r1−23+3r2−23+3r3−23\sqrt[3]{3 r_{1}-2}+\sqrt[3]{3 r_{2}-2}+\sqrt[3]{3 r_{3}-2}.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Let rr be a root of the given polynomial. Then r3−3r2+1=0⟹r3−3r2+3r−1=3r−2⟹r−1=3r−23r^{3}-3 r^{2}+1=0 \Longrightarrow r^{3}-3 r^{2}+3 r-1=3 r-2 \Longrightarrow r-1=\sqrt[3]{3 r-2} Now by Vieta the desired value is r1+r2+r3−3=3−3=0r_{1}+r_{2}+r_{3}-3=3-3=0.

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